Find the Inverse f(x)=(2x+1)/(3x)

$f (x) = \frac{2 x + 1}{3 x}$

Write $f (x) = \frac{2 x + 1}{3 x}$ as an equation.

$y = \frac{2 x + 1}{3 x}$

Interchange the variables.

$x = \frac{2 y + 1}{3 y}$

Solve for $y$ .

Rewrite the equation as $\frac{2 y + 1}{3 y} = x$ .

$\frac{2 y + 1}{3 y} = x$

Find the LCD of the terms in the equation.

Finding the LCD of a list of values is the same as finding the LCM of the denominators of those values.

$3 y, 1$

The LCM of one and any expression is the expression.

$3 y$

Multiply each term in $\frac{2 y + 1}{3 y} = x$ by $3 y$ to eliminate the fractions.

Multiply each term in $\frac{2 y + 1}{3 y} = x$ by $3 y$ .

$\frac{2 y + 1}{3 y} (3 y) = x (3 y)$

Simplify the left side.

Rewrite using the commutative property of multiplication.

$3 \frac{2 y + 1}{3 y} y = x (3 y)$

Cancel the common factor of $3$ .

Factor $3$ out of $3 y$ .

$3 \frac{2 y + 1}{3 (y)} y = x (3 y)$

Cancel the common factor.

$3 \frac{2 y + 1}{3 y} y = x (3 y)$

Rewrite the expression.

$\frac{2 y + 1}{y} y = x (3 y)$

Cancel the common factor of $y$ .

Cancel the common factor.

$\frac{2 y + 1}{y} y = x (3 y)$

Rewrite the expression.

$2 y + 1 = x (3 y)$

Simplify the right side.

Rewrite using the commutative property of multiplication.

$2 y + 1 = 3 x y$

Solve the equation.

Subtract $3 x y$ from both sides of the equation.

$2 y + 1 - 3 x y = 0$

Subtract $1$ from both sides of the equation.

$2 y - 3 x y = - 1$

Factor $y$ out of $2 y - 3 x y$ .

Factor $y$ out of $2 y$ .

$y \cdot 2 - 3 x y = - 1$

Factor $y$ out of $- 3 x y$ .

$y \cdot 2 + y (- 3 x) = - 1$

Factor $y$ out of $y \cdot 2 + y (- 3 x)$ .

$y (2 - 3 x) = - 1$

Divide each term in $y (2 - 3 x) = - 1$ by $2 - 3 x$ and simplify.

Divide each term in $y (2 - 3 x) = - 1$ by $2 - 3 x$ .

$\frac{y (2 - 3 x)}{2 - 3 x} = \frac{- 1}{2 - 3 x}$

Simplify the left side.

Cancel the common factor of $2 - 3 x$ .

Cancel the common factor.

$\frac{y (2 - 3 x)}{2 - 3 x} = \frac{- 1}{2 - 3 x}$

Divide $y$ by $1$ .

$y = \frac{- 1}{2 - 3 x}$

Simplify the right side.

Move the negative in front of the fraction.

$y = - \frac{1}{2 - 3 x}$

Replace $y$ with $f^{- 1} (x)$ to show the final answer.

$f^{- 1} (x) = - \frac{1}{2 - 3 x}$

Verify if $f^{- 1} (x) = - \frac{1}{2 - 3 x}$ is the inverse of $f (x) = \frac{2 x + 1}{3 x}$ .

To verify the inverse, check if $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ .

Evaluate $f^{- 1} (f (x))$ .

Set up the composite result function.

$f^{- 1} (f (x))$

Evaluate $f^{- 1} (\frac{2 x + 1}{3 x})$ by substituting in the value of $f$ into $f^{- 1}$ .

$f^{- 1} (\frac{2 x + 1}{3 x}) = - \frac{1}{2 - 3 \frac{2 x + 1}{3 x}}$

Simplify the denominator.

Cancel the common factor of $3$ .

Factor $3$ out of $- 3$ .

$f^{- 1} (\frac{2 x + 1}{3 x}) = - \frac{1}{2 + 3 (- 1) (\frac{2 x + 1}{3 x})}$

Factor $3$ out of $3 x$ .

$f^{- 1} (\frac{2 x + 1}{3 x}) = - \frac{1}{2 + 3 (- 1) (\frac{2 x + 1}{3 (x)})}$

Cancel the common factor.

$f^{- 1} (\frac{2 x + 1}{3 x}) = - \frac{1}{2 + 3 \cdot (- 1 \frac{2 x + 1}{3 x})}$

Rewrite the expression.

$f^{- 1} (\frac{2 x + 1}{3 x}) = - \frac{1}{2 - 1 \frac{2 x + 1}{x}}$

Rewrite $- 1 \frac{2 x + 1}{x}$ as $- \frac{2 x + 1}{x}$ .

$f^{- 1} (\frac{2 x + 1}{3 x}) = - \frac{1}{2 - \frac{2 x + 1}{x}}$

To write $2$ as a fraction with a common denominator, multiply by $\frac{x}{x}$ .

$f^{- 1} (\frac{2 x + 1}{3 x}) = - \frac{1}{\frac{2 x}{x} - \frac{2 x + 1}{x}}$

Combine the numerators over the common denominator.

$f^{- 1} (\frac{2 x + 1}{3 x}) = - \frac{1}{\frac{2 x - (2 x + 1)}{x}}$

Rewrite $\frac{2 x - (2 x + 1)}{x}$ in a factored form.

Apply the distributive property.

$f^{- 1} (\frac{2 x + 1}{3 x}) = - \frac{1}{\frac{2 x - (2 x) - 1 \cdot 1}{x}}$

Multiply $2$ by $- 1$ .

$f^{- 1} (\frac{2 x + 1}{3 x}) = - \frac{1}{\frac{2 x - 2 x - 1 \cdot 1}{x}}$

Multiply $- 1$ by $1$ .

$f^{- 1} (\frac{2 x + 1}{3 x}) = - \frac{1}{\frac{2 x - 2 x - 1}{x}}$

Subtract $2 x$ from $2 x$ .

$f^{- 1} (\frac{2 x + 1}{3 x}) = - \frac{1}{\frac{0 - 1}{x}}$

Subtract $1$ from $0$ .

$f^{- 1} (\frac{2 x + 1}{3 x}) = - \frac{1}{\frac{- 1}{x}}$

Move the negative in front of the fraction.

$f^{- 1} (\frac{2 x + 1}{3 x}) = - \frac{1}{- \frac{1}{x}}$

Cancel the common factor of $1$ and $- 1$ .

Rewrite $1$ as $- 1 (- 1)$ .

$f^{- 1} (\frac{2 x + 1}{3 x}) = - \frac{- 1 \cdot - 1}{- \frac{1}{x}}$

Move the negative in front of the fraction.

$f^{- 1} (\frac{2 x + 1}{3 x}) = \frac{1}{\frac{1}{x}}$

Multiply the numerator by the reciprocal of the denominator.

$f^{- 1} (\frac{2 x + 1}{3 x}) = 1 x$

Multiply $x$ by $1$ .

$f^{- 1} (\frac{2 x + 1}{3 x}) = x$

Evaluate $f (f^{- 1} (x))$ .

Set up the composite result function.

$f (f^{- 1} (x))$

Evaluate $f (- \frac{1}{2 - 3 x})$ by substituting in the value of $f^{- 1}$ into $f$ .

$f (- \frac{1}{2 - 3 x}) = \frac{2 (- \frac{1}{2 - 3 x}) + 1}{3 (- \frac{1}{2 - 3 x})}$

Simplify the numerator.

Multiply $2 (- \frac{1}{2 - 3 x})$ .

Multiply $- 1$ by $2$ .

$f (- \frac{1}{2 - 3 x}) = \frac{- 2 \frac{1}{2 - 3 x} + 1}{3 (- \frac{1}{2 - 3 x})}$

Combine $- 2$ and $\frac{1}{2 - 3 x}$ .

$f (- \frac{1}{2 - 3 x}) = \frac{\frac{- 2}{2 - 3 x} + 1}{3 (- \frac{1}{2 - 3 x})}$

Move the negative in front of the fraction.

$f (- \frac{1}{2 - 3 x}) = \frac{- \frac{2}{2 - 3 x} + 1}{3 (- \frac{1}{2 - 3 x})}$

Write $1$ as a fraction with a common denominator.

$f (- \frac{1}{2 - 3 x}) = \frac{- \frac{2}{2 - 3 x} + \frac{2 - 3 x}{2 - 3 x}}{3 (- \frac{1}{2 - 3 x})}$

Combine the numerators over the common denominator.

$f (- \frac{1}{2 - 3 x}) = \frac{\frac{- 2 + 2 - 3 x}{2 - 3 x}}{3 (- \frac{1}{2 - 3 x})}$

Reorder terms.

$f (- \frac{1}{2 - 3 x}) = \frac{\frac{- 3 x - 2 + 2}{- 3 x + 2}}{3 (- \frac{1}{2 - 3 x})}$

Rewrite $\frac{- 3 x - 2 + 2}{- 3 x + 2}$ in a factored form.

Add $- 2$ and $2$ .

$f (- \frac{1}{2 - 3 x}) = \frac{\frac{- 3 x + 0}{- 3 x + 2}}{3 (- \frac{1}{2 - 3 x})}$

Add $- 3 x$ and $0$ .

$f (- \frac{1}{2 - 3 x}) = \frac{\frac{- 3 x}{- 3 x + 2}}{3 (- \frac{1}{2 - 3 x})}$

Move the negative in front of the fraction.

$f (- \frac{1}{2 - 3 x}) = \frac{- \frac{3 x}{- 3 x + 2}}{3 (- \frac{1}{2 - 3 x})}$

Simplify the denominator.

Multiply $- 1$ by $3$ .

$f (- \frac{1}{2 - 3 x}) = \frac{- \frac{3 x}{- 3 x + 2}}{- 3 \frac{1}{2 - 3 x}}$

Combine $- 3$ and $\frac{1}{2 - 3 x}$ .

$f (- \frac{1}{2 - 3 x}) = \frac{- \frac{3 x}{- 3 x + 2}}{\frac{- 3}{2 - 3 x}}$

Reduce the expression by cancelling the common factors.

Move the negative in front of the fraction.

$f (- \frac{1}{2 - 3 x}) = \frac{- \frac{3 x}{- 3 x + 2}}{- \frac{3}{2 - 3 x}}$

Dividing two negative values results in a positive value.

$f (- \frac{1}{2 - 3 x}) = \frac{\frac{3 x}{- 3 x + 2}}{\frac{3}{2 - 3 x}}$

Multiply the numerator by the reciprocal of the denominator.

$f (- \frac{1}{2 - 3 x}) = \frac{3 x}{- 3 x + 2} \cdot \frac{2 - 3 x}{3}$

Cancel the common factor of $3$ .

Factor $3$ out of $3 x$ .

$f (- \frac{1}{2 - 3 x}) = \frac{3 (x)}{- 3 x + 2} \cdot \frac{2 - 3 x}{3}$

Cancel the common factor.

$f (- \frac{1}{2 - 3 x}) = \frac{3 x}{- 3 x + 2} \cdot \frac{2 - 3 x}{3}$

Rewrite the expression.

$f (- \frac{1}{2 - 3 x}) = \frac{x}{- 3 x + 2} \cdot (2 - 3 x)$

Multiply $\frac{x}{- 3 x + 2}$ by $2 - 3 x$ .

$f (- \frac{1}{2 - 3 x}) = \frac{x (2 - 3 x)}{- 3 x + 2}$

Cancel the common factor of $2 - 3 x$ and $- 3 x + 2$ .

Reorder terms.

$f (- \frac{1}{2 - 3 x}) = \frac{x (- 3 x + 2)}{- 3 x + 2}$

Cancel the common factor.

$f (- \frac{1}{2 - 3 x}) = \frac{x (- 3 x + 2)}{- 3 x + 2}$

Divide $x$ by $1$ .

$f (- \frac{1}{2 - 3 x}) = x$

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = - \frac{1}{2 - 3 x}$ is the inverse of $f (x) = \frac{2 x + 1}{3 x}$ .

$f^{- 1} (x) = - \frac{1}{2 - 3 x}$

Find the Inverse f(x)=(2x+1)/(3x)

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